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  • UVa 10269 Adventure of Super Mario (Floyd + DP + BFS)

    题意:有A个村庄,B个城市,m条边,从起点到终点,找一条最短路径。但是,有一种工具可以使人不费力的移动L个长度,但始末点必须是城市或村庄。这种工具有k个,每个只能使用一次,并且在城市内部不可使用,但在村庄内部可使用。另外,在城市或村庄内部的时间不计。

    析:先预处理出来使用工具能到达的距离,这个可以用Floyd 来解决,然后f[i][u] 表示到达 u 还剩下 i 次工具未用,然后用bfs就可以很简单的解决。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #include <list>
    #include <assert.h>
    #include <bitset>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define fi first
    #define se second
    #define pb push_back
    #define sqr(x) ((x)*(x))
    #define ms(a,b) memset(a, b, sizeof a)
    #define sz size()
    #define pu push_up
    #define pd push_down
    #define cl clear()
    #define all 1,n,1
    #define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 1e20;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 100 + 10;
    const int maxm = 1e5 + 10;
    const int mod = 50007;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, -1, 0, 1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c) {
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    
    int dp[maxn][maxn];
    int f[12][maxn];
    
    int main(){
      int T;  cin >> T;
      while(T--){
        int A, B, L, K;
        scanf("%d %d %d %d %d", &A, &B, &m, &L, &K);
        ms(dp, INF);
        while(m--){
          int u, v, c;  scanf("%d %d %d", &u, &v, &c);
          dp[u][v] = dp[v][u] = min(c, dp[u][v]);
        }
        FOR(k, 1, A+1)  FOR(i, 1, A+B+1)  FOR(j, 1, A+B+1)
          dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
        ms(f, INF);  f[K][A+B] = 0;
        queue<P> q;  q.push(P(K, A+B));
        while(!q.empty()){
          P p = q.front();  q.pop();
          int u = p.se, i = p.fi;
          for(int v = 1; v < A+B; ++v){
            if(dp[u][v] == INF || u == v)  continue;
            if(f[i][v] > f[i][u] + dp[u][v]){
              f[i][v] = f[i][u] + dp[u][v];
              q.push(P(i, v));
            }
            if(i && f[i-1][v] > f[i][u] && L >= dp[u][v]){
              f[i-1][v] = f[i][u];
              q.push(P(i-1, v));
            }
          }
        }
        int ans = INF;
        for(int i = 0; i <= K; ++i)  ans = min(ans, f[i][1]);
        printf("%d
    ", ans);
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7742618.html
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